# stress/strain in my torque wrench--please check

I was checking out my torque wrench. I measure it to be 429mm between the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm. The indicator is a little short of the outer pivot---the deflection at the outer pivot for 150lbf is around 41mm.
I found the formula for deflection: delta = F L^3 / (3 E I)
E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel) F is force applied, L the beam length I is the areal moment of inertia, I = pi/2 r^4
I am going to use the torque instead of force, T=FL, i.e.
delta = T L^2 / (3 E I) = 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)
'units' calculates that as 30mm, a little too short of 41mm measured. Am I making a stupid error somewhere?
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On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski

First thing I'd look at is the assumption (implied by the 3EI term) that the wrench acts like an ideal cantilever.
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Ned Simmons

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On 9/2/2014 11:30 PM, Ned Simmons wrote:

There's also room for error in the assumption for E, since you can't be sure what sort of steel was used for the beam.
If you doubt the accuracy of your wrench, clamp it in a vise and hang some weights off it. It's better to know how it actually responds to a load rather than how it "should" respond.
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On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski

First, your value for Young's modulus is the one for stainless steel. Assuming it's any alloy other than a high-nickel grade, try 210. Any grade of steel will be in between those two figures, 180 and 210, but 210 is probably very close for any steel grade used for the beam of a torque wrench.
I am not going to do the calculations. <g> However, your formula for area moment of inertia doesn't look right. I don't have my statics book handy but I think it should be I = pi/4*r^4.
Good luck.
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Ed Huntress

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On Tue, 02 Sep 2014 23:52:35 -0400, Ed Huntress wrote:

OK, thanks---it's chrome-plated rusty-spotted rod so you're probably right. The change from that is in the wrong direction, though :)

Here's a great tip: just drop this expression in Google---yes, the Google search field---
150 lbf ft * (429mm)^2/(3 * 180 GPa * (pi/2)* (12.4mm/2)^4) , in mm
and good old G responds with 29.8625457 millimeters---COOL, HEH?
I got lazy and started dropping the multiplications ('*') because 'units' allows me to, but Google requires at least some to recognize arithmetical expressions

OK, you got me there--you're good. I think I inadvertently used Iz. Still, Google sez 150 lbf ft (429mm)^2/(3 * 210 GPa (pi/4) (12.4mm/2)^4) , in mm
is 51.1929355 millimeters. Now it's too much :(
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On Thu, 4 Sep 2014 21:28:48 +0000 (UTC), Przemek Klosowski

I hoped it would all balance out with the pi/4 factor, but I didn't try to do it in my head, so I really didn't know which way it would go.

OOH! I LIKE that! Thanks for the tip!
I'll get some use out of that.

Yeah, backing into it, it takes a value of 260 to make it work, and that's just not possible with any steel. In general, the Young's modulus for steel falls into a very narrow range, except for high-nickel alloys (including stainless), which are lower, and high-speed steel with tungsten, which is higher. Your torque-wrench beam is most likely a plain-carbon grade, which should be in the range of 205 - 210.
Assuming your measurements are right, I don't know where else to go with it.
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Ed Huntress

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On Thu, 04 Sep 2014 18:09:58 -0400, Ed Huntress wrote:

Given that the dependence is on diameter^4, it wouldn't take much of a measurement error to throw it off.
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Tim Wescott
Wescott Design Services
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On Thu, 04 Sep 2014 18:18:32 -0500, Tim Wescott

Yeah, that's what I was thinking, too.
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Ed Huntress